poj1159 Palindrome dp에 최소 문자를 추가하여 메모를 구성합니다.

Palindrome
Time Limit: 3000MS
 
Memory Limit: 65536K
Total Submissions: 44186
 
Accepted: 15050
Description
A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome.
As an example, by inserting 2 characters, the string "Ab3bd"can be transformed into a palindrome ("dAb3bAd"or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome.
Input
Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters from 'A' to 'Z', lowercase letters from 'a' to 'z' and digits from '0' to '9'. Uppercase and lowercase letters are to be considered distinct.
Output
Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.
Sample Input

5
Ab3bd

Sample Output

2

Source
IOI 2000
 
이러한 결론에 따라 추가할 문자의 개수 = 원래 문자열의 길이 - 원래 문자열과 역 문자열의 가장 긴 공통 서열의 길이.
그리고 스크롤 그룹을 사용하면 됩니다.

#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<memory.h>
#include<string.h>
using namespace std;
char str[5010];
char s2[5010];
int dp[2][5010];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        scanf("%s",str);
        for(int i=n-1;i>=0;i--)
        s2[n-i-1]=str[i];
        s2[n]='\0';
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            if(str[i-1]==s2[j-1])
            dp[i%2][j]=dp[(i-1)%2][j-1]+1;
            else
            dp[i%2][j]=max(dp[(i-1)%2][j],dp[i%2][j-1]);
        }
        int l=dp[n%2][n];
        cout<<n-l<<endl;
    }
}